Proving the Monotonicity of Outer Measures: A Simple Task or a Tricky Proof?

[MrGandalf]

I have a sneaky suspicion that this is a fairly simple task, but I just can't seem to break through this particular proof.

Homework Statement

The (Lebesgue) outer measure of any set A\subseteq\mathbb{R} is:

m^*(A) = inf Z_A

where

Z_A = \bigg\{\sum_{n=1}^\infty l(I_n)\;:\;I_n\;\text{are intervals},\;A\subseteq\bigcup_{n=1}^\infty I_n\bigg\}

My problem is to prove that m^* is monotone, i.e

If A\subset B then m^*(A) \leq m^*(B)

The hints are to show that Z_B \subset Z_A and then use the definition of the infimum to show that the larger set can't have an infimum greater than the smaller set.

The attempt at a solution

If I_n covers B, then it will also cover A.
A\subset B\subset \bigcup_n I_n. Hence Z_B \subset Z_A.

And now I am confused! :) I have two questions.

1) I was under the impression that l(I_n) was the length of a certain intervall. But that makes Z_A and Z_B numbers. How can one number be a subset of another number?

2) My other question is regarding the statement about infimums. What does the size of a set have to do with what the infimum can be? For instance.
A = \{2, 3\} and B = \{1,2,3,4\}. Here A\subset B but \text{inf}\; B < \text{inf}\; A

Hopefully someone can shed some light on this!

 

[AKG]

1) No, Z_A and Z_B are sets of numbers. For each open interval covering (I_n) of A, you get the number \sum l(I_n). As you range over all such coverings, you get a whole bunch of different numbers, and Z_A is the set of those numbers.

2) The size of a set doesn't affect its inf, but if A is a subset of B, then B's inf is less than or equal to A's. Because inf(B) is the greatest lower bound of B, and if A is a subset of B, then every lower bound of B is a lower bound of A, and hence the greatest lower bound of B is a lower bound for A, and hence less than or equal to the greatest lower bound for A.

 

[MrGandalf]

Thank you!

I will charge the proof with newfound courage and motivation. :)